ELEC 7770 Advanced VLSI Design Spring 2007 Constraint Graph and Performance Optimization

Spring 07, Apr 10, 12Spring 07, Apr 10, 12 ELEC 7770: Advanced VLSI Design (Agrawal)ELEC 7770: Advanced VLSI Design (Agrawal) 11

ELEC 7770ELEC 7770Advanced VLSI DesignAdvanced VLSI Design

Spring 2007Spring 2007Constraint Graph and Performance Constraint Graph and Performance

OptimizationOptimization

Vishwani D. AgrawalVishwani D. AgrawalJames J. Danaher ProfessorJames J. Danaher Professor

ECE Department, Auburn UniversityECE Department, Auburn University

Auburn, AL 36849Auburn, AL 36849

[email protected]@eng.auburn.edu

http://www.eng.auburn.edu/~vagrawal/COURSE/E7770_Spr07http://www.eng.auburn.edu/~vagrawal/COURSE/E7770_Spr07


Retiming TheoremRetiming Theorem Given a network G(V, E, W) and a cycle time T, Given a network G(V, E, W) and a cycle time T,

(r1, . . . ) is a feasible retiming if and only if:(r1, . . . ) is a feasible retiming if and only if: ri – rj ri – rj ≤ wij≤ wij for all edges (vi,vj) for all edges (vi,vj) εε E E ri – rj ≤ W(vi,vj) – 1 ri – rj ≤ W(vi,vj) – 1 for all node-pairs vi, vj such thatfor all node-pairs vi, vj such that

D(vi,vj) D(vi,vj) > T> T

Where,Where,

W(vi,vj) is the minimum weight path between vi and vjW(vi,vj) is the minimum weight path between vi and vj

D(vi,vj) is the maximum delay among all minimum D(vi,vj) is the maximum delay among all minimum weight paths between vi and vjweight paths between vi and vj


Timing OptimizationTiming Optimization

Find the clock period (T) by path analysis.Find the clock period (T) by path analysis. Set clock period to T/2 and find a feasible Set clock period to T/2 and find a feasible

retiming.retiming. If feasible, further reduce the clock period to If feasible, further reduce the clock period to

half.half. If not feasible, increase clock period.If not feasible, increase clock period. Do a binary search for optimum clock period.Do a binary search for optimum clock period. Retime the circuit.Retime the circuit.


Representing a ConstraintRepresenting a Constraint

ri – rj ≤ wij or rj ≥ ri – wij

rj ri– wij


Constraint GraphConstraint Graph

r1 ≥ r0 + 3 r1 ≥ r2 + 1 r2 ≥ r0 + 1 r2 ≥ r1 – 1 r3 ≥ r1 + 1 r3 ≥ r2 + 4 r0 ≥ r3 – 6 r0

r1

r2

r3-1 1

3 1

1 4

-6


Feasibility ConditionFeasibility Condition

A set of values for variables can be found if and A set of values for variables can be found if and only if the constraint graph has no positive only if the constraint graph has no positive cycles.cycles.

This is also the condition for the solvability of the This is also the condition for the solvability of the longest path problem, which provides a solution longest path problem, which provides a solution to the set of constraints.to the set of constraints.


Example: Infeasible ConstraintsExample: Infeasible Constraints

x1 ≥ x2 + 6 x2 ≥ x1 – 3

x1 x2

6

-3

x1

x2

60x1 ≥ x2 + 6

x2 ≥ x1 – 3

3

3

Positive cycle mean no longest path can be found.


Solving a Constraint SetSolving a Constraint Set

r1 ≥ r0 + 3 r1 ≥ r2 + 1 r2 ≥ r0 + 1 r2 ≥ r1 – 1 r3 ≥ r1 + 1 r3 ≥ r2 + 4 r0 ≥ r3 – 6 r0

r1

r2

r3-1 1

3 1

1 4

-6

Longest path from source r0: r0, r1, r2, r3Path lengths: s0=0, s1=3, s2=2, s3=6Solution: r0=0, r1=3, r2=2, r3=6


The General Path ProblemThe General Path Problem

Find the shortest (or longest) path in a graph Find the shortest (or longest) path in a graph from a source vertex to any other vertex.from a source vertex to any other vertex.

Graph has vertices and directed edges:Graph has vertices and directed edges: Edge weights can be positive or negativeEdge weights can be positive or negative Graph can be cyclicGraph can be cyclic Single source vertex – a vertex with 0 in-degreeSingle source vertex – a vertex with 0 in-degree

Inconsistent problemInconsistent problem Negative cycles for shortest pathNegative cycles for shortest path Positive cycles for longest pathPositive cycles for longest path


Dijkstra’s Shortest Path AlgorithmDijkstra’s Shortest Path Algorithm

Greedy algorithm.Greedy algorithm. Applies to directed acyclic graphs (DAG) with Applies to directed acyclic graphs (DAG) with positivepositive

edge weights.edge weights. Computational complexityComputational complexity

O(|E| + |V| log |V|) O(|E| + |V| log |V|) ≤ O(n≤ O(n22)) References:References:

A. Aho, J. Hopcroft and J. Ullman, A. Aho, J. Hopcroft and J. Ullman, Data Structures and Data Structures and AlgorithmsAlgorithms, Reading, Massachusetts: Addison-Wesley, 1983., Reading, Massachusetts: Addison-Wesley, 1983.

T. Cormen, C. Leiserson and R. Rivest, T. Cormen, C. Leiserson and R. Rivest, Introduction to Introduction to AlgorithmsAlgorithms, New York: McGraw-Hill, 1990., New York: McGraw-Hill, 1990.


Dijkstra’s Shortest Path AlgorithmDijkstra’s Shortest Path Algorithm

v0

v2

v3

v1w01=15 3

10

2 6source

si = path weight (v0, vi)

Alg. stepsAlg. steps s0s0 s1s1 s2s2 s3s3

Initially: mark s0Initially: mark s0 00 1515 22

Step 1: mark s2Step 1: mark s2 00 1212 22 88



Each step marks the path with smallest weight and updates the unmarked path weights.


Dijkstra’s Algorithm, G(V, E, W)Dijkstra’s Algorithm, G(V, E, W)

s0(1) = 0s0(1) = 0 initialize sourceinitialize source

for ( i = 1 to n )for ( i = 1 to n ) initialize path weights, n=|V| –1initialize path weights, n=|V| –1si(1) = w0isi(1) = w0i

repeat {repeat {

Select an unmarked vertex vq such that sq is minimalSelect an unmarked vertex vq such that sq is minimal

Mark vqMark vq

foreach ( unmarked vertex vi )foreach ( unmarked vertex vi )si =si = min min { si, sq + wqi } { si, sq + wqi }

}}until (all vertices are marked)until (all vertices are marked)


Dijkstra’s Longest Path AlgorithmDijkstra’s Longest Path Algorithm

v0

v2

v3

v1w01=15 3

10

2 6source

si = path length (v0, vi)


InitiallyInitially 00 -15-15 -2-2

Step 1Step 1 00 -15-15 -2-2

Step 2Step 2 00 -15-15 -2-2 -8-8

Step 2Step 2 00 -15-15 -2-2 -8-8v0

v2

v3

v1w01= -15 -3

-10

-2 -6source


Dijkstra’s Alg. for Cycles, Neg. WeightsDijkstra’s Alg. for Cycles, Neg. Weights

v0

v2

v3

v1w01=15 3

5

2 4source



InitiallyInitially 00 1515 22

Step 1Step 1 00 77 22 66

Step 2Step 2 00 77 22 66

Step 3Step 3 00 77 22 6?6?

-2

There exists a v0 to v3 path of length 5


Bellman’s Equations – Shortest PathBellman’s Equations – Shortest Path

vi

vn

vm

vkvj

sq = minimum path weight betweensource and vq

wki

wji

wmi

wni

For all vertices:

si = min (sq + wqi)

vq ε pred(vi)


Bellman-Ford Algorithm, G(V, E, W)Bellman-Ford Algorithm, G(V, E, W)Bellman-Ford {Bellman-Ford {

s0(1) = 0s0(1) = 0 initialize sourceinitialize source

for ( i = 1 to n )for ( i = 1 to n ) initialize path weights, n = |V| – 1initialize path weights, n = |V| – 1si(1) = w0isi(1) = w0i

for ( j = 1 to n )for ( j = 1 to n ) n iterationsn iterationsfor ( i = 1 to n )for ( i = 1 to n )

si(j+1) =si(j+1) = min min { si(j), sk(j) + wkj } { si(j), sk(j) + wkj }

vvk k εε pred(vi) pred(vi)

}}

if ( si(j+1) == si(j) if ( si(j+1) == si(j) i ) return (true)i ) return (true)

}}

return (false)return (false) Complexity = O(|V||E|) ≤ O(n3)


Bellman-Ford Shortest PathBellman-Ford Shortest Path

v0

v2

v3

v1w01=15 3

10

2 6source




Iteration 1Iteration 1 00 1212 22 88



n = 3


Bellman-Ford Longest PathBellman-Ford Longest Path

v0

v2

v3

v1w01= -15 -3

-10

-2 -6source



InitiallyInitially 00 -15-15 -2-2

Iteration 1Iteration 1 00 -15-15 -2-2 -8-8

Iteration 2Iteration 2 00 -15-15 -2-2 -8-8

n = 3 (shortest path)

Reverse the sign of weights and solve shortest path problem.(Alternative: keep original weights and change min operator in algorithm to max.)

Weights reversed


Bellman’s Equations – Longest PathBellman’s Equations – Longest Path

vi

vn

vm

vkvj

sq = maximum path weight betweensource and vq

wki

wji

wmi

wni

For all vertices:

si = max (sq + wqi)

vq ε pred(vi)


Bellman-Ford for Cycles, Neg. WeightsBellman-Ford for Cycles, Neg. Weights

v0

v2

v3

v1w01=15 3

5

2 4source







-2 n = 3 (shortest path)

This was incorrect with Dijkstra’s shortest path algorithm


Bellman-Ford for Negative CycleBellman-Ford for Negative Cycle

v0

v2

v3

v1w01=15 -3

5

2 4source







2

Values not stabilized after n iterations.Inconsistent problem: negative cycle.

n = 3 (shortest path)


Retiming ExampleRetiming Example

FF10 5 5

Delay

a b c


Retiming GraphRetiming Graph

FF10 5 5a b c

h0

a10

b5

c5

0 0 1

1

Critical path = 15It is the longest path consisting only of zero weight edges.


Feasibility ConstraintsFeasibility Constraints

FF10 5 5a b c

h0

a10

b5

c5

0 0 1

1

ri – rj ≤ wij edges i → jRetiming should not cause negative edge weights.

rh – ra ≤ 0ra – rb ≤ 0rb – rc ≤ 1rc – rh ≤ 1


Constraint GraphConstraint Graph

FF10 5 5a b c

rh0

ra10

rb5

rc5

0 0 -1

-1

ri – rj ≤ wij edges i → jRetiming should not cause negative edge weights.

rh – ra ≤ 0ra – rb ≤ 0rb – rc ≤ 1rc – rh ≤ 1

Observation: Constraint graph has the same structure as the original retiming graph, with signs of weights reversed. Vertex labels are the retiming integer variables.


Max Delay for Min Weight PathsMax Delay for Min Weight Paths

h0

a10

b5

c5

0 0 1

1

W(h,a) = 0 D(h,a) = 10W(h,b) = 0 D(h,b) = 15W(h,c) = 1 D(h,c) = 20W(a,b) = 0 D(a,b) = 15W(a,c) = 1 D(a,c) = 20W(a,h) = 2 D(a,h) = 20

W(b,c) = 1 D(b,c) = 10W(b,h) = 2 D(b,h) = 10W(b,a) = 2 D(b,a) = 20W(c,h) = 1 D(c,h) = 5W(c,a) = 1 D(c,a) = 15W(c,b) = 1 D(c,b) = 20

T = 15


Timing Optimization, T = 7.5?Timing Optimization, T = 7.5?



rh0

ra10

rb5

rc5

0 0 -1

-1

ri – rj ≤ W(I,j) – 1 paths (i,j) such that D(i,j) > 7.5

Constraint graph(feasibility)





rh0

ra10

rb5

rc5

0 0 -1

-1

11

0

1

0

-1 0

-1

-1

0

0

Positive cycleNo solution





rh0

ra10

rb5

rc5

0 0 -1

-1

10

1

0

-1 -1

0

0

rh = 0 rb = 1 rc = 0 ra = 0


Retiming GraphRetiming Graph

FF10 5 5a b c

h0

a10

b5

c5

0 0 1

1

rh = 0 ra = 0 rb = 1 rc = 0

1 0

wij_retimed = wij + rj – ri


Retimed CircuitRetimed Circuit

FF10 5 5a b c

h0

a10

b5

c5

0

1

rh = 0 ra = 0 rb = 1 rc = 0

1 0

Critical Path = 10

Logic optimization will remove these.

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ELEC 7770 Advanced VLSI Design Spring 2007 Constraint Graph and Performance Optimization